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# PHYS 200: Fundamentals of Physics I

## Lecture 5

## - Work-Energy Theorem and Law of Conservation of Energy

### Overview

The lecture begins with a review of the loop-the-loop problem. Professor Shankar then reviews basic terminology in relation to work, kinetic energy and potential energy. He then goes on to define the Work-Energy Theorem. Finally, the Law of Conservation of Energy is discussed and demonstrated with specific examples.

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html## Fundamentals of Physics I## PHYS 200 - Lecture 5 - Work-Energy Theorem and Law of Conservation of Energy## Chapter 1. More on Loop-the-Loop and Intro to Concept of Energy [00:00:00]
So yes, this will have an acceleration, you cannot avoid it. mg acting, it’s accelerating too. What’s the difference between these two? Here, we start with zero velocity; you accelerate downwards. That means you pick up a downward velocity. As it accelerates you keep falling down and picking up more and more speed. The same acceleration’s experienced by the cart also, and a little more because T is also pushing down. But there the change in velocity over a tiny amount of time is not added to zero velocity, but to this large velocity it has at that instant. And the tiny increment in velocity if added this way will produce a velocity in that direction, and what that really means if you wait a small amount of time, this thing is going along the new velocity direction.So, that’s the difference between–So, the acceleration is the same for a falling apple as for the loop-the-loop cart. But in the apple’s case, it is added to zero initial velocity so the acceleration and the velocity are the same, namely down. If you have a huge velocity to which you add a tiny change in velocity, the downward direction you rotate the velocity vector that just means that as it goes around the circle, the velocity vector is constantly tangent to the circle. There’s another way in which people understand this. If you are on the Earth and you are standing, say, on a building and you fire a gun, it goes like that and it hits the ground because of acceleration due to gravity. If you fire it a little faster it’ll go there. There’ll be a certain speed at which it keeps falling but doesn’t get any closer to the center. That is in fact how you launch a bullet into orbit. So, what’s the first thing you should do when you fire this gun? Move away because it’s going to come back and get you from behind, okay [students laughing]. Alright that’s a very useful lesson. So, you might think, “Hey I shot it in that direction, it’s not my problem.” Like everything else it’ll come back and get you when you least expect it. This is how you have–;So, is this thing falling or not? It is constantly falling but the Earth is falling under it so it’s endless search to come to the Earth. Whereas if you drop the bullet, it’ll just fall and it’ll hit the ground; that’s what we always think as drop. Okay, so this is something I did in a rush, and I didn’t want that because this is really the part of physics which is not intuitive, and I think we all know what the main point is. The main point is velocity is a vector and its value can change due to change in direction, not change in magnitude. Yes sir?
v, put the r, get some number, multiply by the Δt, 1/10^{th} of a second. That’ll be the change in that time. So, in that time also the guy would have moved over in the circle a little bit, just precisely so that at that point your new velocity vector is tangent to your new location. So this little Δv I drew is not a fixed number; it depends on how long you want to wait. Calculus says wait an infinitesimal amount of time, in practice wait 1/10^{th} of a second and it’ll be fine.Okay, today there is a very, very important concept I want to introduce. It’s a very robust and powerful concept; it has to do with energy. It is powerful because after the laws of quantum mechanics were discovered in the sub atomic world, we basically gave up on many cherished notions. You know, you must have heard that particles cannot have a definite position and definite velocity at a given time. They don’t have any trajectories. You see a particle here now, then you see it there. That’s true because they’re observed events. What happens in between? You might think they have a trajectory but they don’t. The forces are not well defined because if you don’t know where the guy is, you don’t know what the velocity is, you don’t know what the acceleration is. So most of the ideas of Newtonian mechanics are basically surrendered, but the notion of energy turns out to be very robust and survives all the quantum revolution. In fact, there is a period when people are studying nuclear reactions and the energy you began with didn’t seem to be the energy you ended up with. So something was missing. So Neils Bohr, father of the atom said, “Maybe the Law of Conservation of Energy is not valid.” Then Pauli said, “I think I put my money on the Law of Conservation of Energy. I would postulate that some other tiny particles that you guys cannot see, that’s carrying away all the missing energy.” That was a very radical thing to do in those days when people did not lightly postulate new particles. Nowadays, if you don’t postulate a particle you don’t get your PhD in particle physics [students laugh]. In those days it was very radical, and the particle was not seen for many, many years. It was called a neutrino. Nowadays, neutrinos are one of the most exciting things one could study. There’s a lot of mysteries in the universe connected with neutrinos. Okay, so energy is a concept you’re going to learn. Here’s again a place where I ask you, don’t memorize the formulas and start plugging them in. Try to follow the logic by which you are driven to this notion, because then it’ll be less luggage for you to carry in your head. And I’ll try to make the notion as natural as possible rather than saying, “This is my definition of energy, this is the definition of kinetic energy.” Let’s ask where does all that come from. How many people have seen kinetic and potential energies in your past? Okay, if you haven’t seen it that’s fine, this is not something you’re supposed to know. If you know it and you learned it properly it’ll be helpful, but you don’t have know that. You just have to know what we have done so far. So let’s go back and ask a simple question. When a force acts on a body, what is its main effect? What does it do? Yes?
So, let us take an example where the force is in the a for a distance d, then the relation between final initial velocities is this. So let’s put that F/m here and let’s put a d.## Chapter 2. Work-Energy Theorem and Power [00:11:57]Since we are interested in the effect of the force, we want to see what the force did to the velocity. What you find here is that if you want to find the change in the velocity, of course you should take this guy to the left-hand side, right? Take it to the other side and you want to isolate the force, so bring m/2v is _{1}^{2}F times d.This is a very important concept. It says, when the force acts on a body it changes the velocity, and it depends on how far the force has been acting, how many meters you’ve been pushing the object. It’s clear that if you didn’t push it at all, then even the forces acting, velocity hasn’t had time to change. So, it seems to depend on how far the force acted, and the change is not simply in velocity but in velocity squared. That’s what comes out; then, you realize that it’s the most natural thing in the world to give this combination a name because that’s what comes out of this. That combination is called kinetic energy, and I’m going to call it K at the end minus K at the beginning is F times d. So F times d is denoted by the symbol W, and that’s called the work done by F. That’s how we introduce a notion of a force times distance as the definition of work. This is how people–you can say, “Why did anyone think of taking the combination F times d?” Well, this is how it happens. So, you need to have a unit for this, which is actually Newton meters, but we are tired of calling everything a Newton meter, so we’re going to call it a joule. So joule or J, is just a shorthand for Newton meters.Alright, so let us now write down this following expression here. Δ Now, what if there are 36 forces acting on the body? Which one should I use? I’m pulling and you’re pushing and…[pause]. It’s got to be the net force because Newton’s law connects the net force to the acceleration. If you and I have a tug of war and we cancel each other out to zero then there’s no acceleration. So, you’ve got to go back to the definition; this is very, very important. This Here’s a body, right, and I’m pushing it this way and you’re pushing it that way, let’s say with equal force. And the body could be at rest or if the body had initial velocity it’ll maintain the initial velocity because velocity is for free. Then, you can ask, “What does this theorem say?” This is called a Work Energy Theorem. The Work Energy Theorem says, “The change in energy is equal to the work done by all the forces.” In this case, there is no change in energy, but there are two forces at work that cancel. Then, it turns out that it is sensible to define the work done by me, which is equal to See, here is a simple example. I have a piece of chalk and I’m lifting it at constant velocity from the ground. Its kinetic energy is not changing. That means the total work done on it is zero, but it’s not because there are no forces on it. There is gravity acting down and there is me countering gravity with exactly the right amount I’m applying the force Alright, so we’ve got this result; let’s do the following. Let us imagine all of this occurs in a time Δ For example, if I climb a 12-story building I’ve done some work; my Okay, now I’m going to do the next generalization, which is when the force is not a constant but varies with
So then, we have to ask what is the Work Energy Theorem when the forces itself vary? So, let’s draw ourselves a force which is varying, here is So, if you eventually went from here to here [distance on graph], let me call that point as x_{2}, then the work done by you is really given by the area under that graph. And in every segment you pick up the change in kinetic energy, you add it all up. When we say add it all up on the left-hand side, all the ΔKs will add up to give me the final minus initial kinetic energy. This one is what is written in the notation of calculus as F(x)dx from x to _{1}x_{2}. This is called the integral of the function F(x).Now again, in a course at this level I’m assuming that you guys know what an integral is. Even if you’ve never heard of an integral, if I give you a function and you never heard of calculus, you can still deal with this problem. Because you’ll come to me and say, “Give me your function.” I’m going to plot it in some kind of graph paper with a grid on it, and I’m just going to count the number of tiny squares enclosed here. That’s the area and that’s the change in kinetic energy. So, integration is finding the area under graphs bounded by the function of the top, Now, a little digression. Even though you’re supposed to know this, I want to make sure–part of my “No Child Left Behind” program, that you all follow this. If you never heard calculus or integral calculus this is going to be a two-minute, three-minute introduction. There is a great secret, which is that if you give me a function and you tell me to find the area under this from x_{2}, I can go to the graph paper and draw that, or there is quite often a trick. You don’t have to draw anything, you know. And I will tell you a little more about what it is. If you give me a function F(x), the area under this graph happens to be the following. There is another function G(x) that will be specified in a moment. And you just have to find G at the upper limit minus G at the lower limit, and that will be the area. And who is this mystery function G? It is any function whose derivative is the given function F.So, the whole business of finding area which was solved around Newton’s time, is if you have a function they want the area under it, you go and think of a function whose derivative is the given function. So, it is the opposite of taking derivatives. Differentiation is an algorithmic process; you take So, if I say G(x) is that function whose derivative is x. So, I know I’ve got to start with ^{3}x to the 4^{th} because when I take derivatives it’ll become an x, but with the 4 in it. So, I fight that by putting a 4 downstairs and that’s the function ^{3}G. Yes?
So, maybe it’s again worth asking for a second, why is this true, you know. Why is F(x)dx from zero to the point x. You agree that as I slide _{2}x back and forth, the area is varying with _{2}x, so it’s legitimate to call that a function of _{2}x. A function is anything that depends on something else, and we all agree that if the point actually moves, the area moves. This is my shorthand for the area. The starting point is nailed at zero; this is _{2}G(x. The left asks what is _{2})G(x. That is the area to this point, that we denoted as _{2}+ Δ)F(x)dx from zero to x, that’s _{2} + ΔG(x._{2}+ Δ)So, what is the difference between the two? The difference between the two is F at the point x times _{2} Δx. In other words, by how much is the area changing when I change the upper limit? Well, it’s changing by the height of the function at the very last moment times the change in x. Well, we know what to do now, divide by Δx and take the limit, and you can see that F(x is _{2})dG/dx at the point x. But the point _{2}x is whatever point you want. So, we drop the label, and we just say that _{2}F as a function of x is dG/dx. This is the origin of the result that the area can be found if you can find the inverse derivative of the function.So, let’s keep a couple of popular examples in mind. One, if ^{n} is x^{n} +1 over n + 1. There are integrals for all kinds of functions, and you check them by taking derivatives and make sure you get back your F.## Chapter 3. Conservation of Energy: K2 + U2 = K1 + U1 [00:29:19]Alright, suppose this is what we learned from the math guys, then go back to the Work Energy Theorem and see what it says, Oh, by the way, I didn’t complete one story; let me finish that. This was, if x, _{2}G(x is the area from zero to the point _{1})x. That is perhaps clear to you, if I only wanted the area between the point _{1}x and the point _{1}x, I can take the area all the way up to _{2}x, subtract from it the area all the way up to _{2}x, and that’s of course the claim that this integral _{1}F(x)dx from x to _{1}x is _{2}G(x. That’s the reason the integral is given by a difference of this _{2}) - G(x_{1})G. So, let me use a shorthand notation and call G(x as simply _{2})G, _{2}G(x is _{1})G. You should understand, _{1}G means _{2}G evaluated at the final point, G to _{1}G evaluated initial point. So, you have got K is _{2} - K_{1}G, but _{2} - G_{1}G is some definite function of x. You’ve got to understand that. For example, it could be x over 2 - _{2}^{2}x over 2 in the problem where _{1}^{2}F is equal to x. Then, let’s rearrange this thing to read K is equal to _{2} - G_{2}K._{1} - G_{1}So, the Work Energy Theorem says that, “If a force is acting on a body, a variable force associated with the force is the function K, it’ll be the same as _{1} - G_{1}K._{2} - G_{2}Now, we don’t like the form in which this is written. We have to make a little cosmetic change, and the cosmetic change is to introduce the function E and you call this combination as _{2}E, we are saying, _{1}E, and that’s called the Conservation of Energy._{1} = E_{2}So, what does conservation mean? Conservation of Energy in physics has a totally different meaning from conservation of energy in daily life. Here it means, when a body’s moving under the effect of this force So, let’s take a simple example. So, we take a rock and we drop it. We know it’s picking up speed; we know it’s losing height. So, you may think maybe there is some combination of height and speed, which does not change in this exchange, and we find the combination by this rule now. In the case of gravity, mgy at the initial time will be equal to ½mv + _{2}^{2}mgy at the final time. And that is the–And the whole thing can be evaluated any time. You take any random time you want. The sum of these two numbers doesn’t change. And that’s a very, very powerful result. So, I’m going to give you one more example later when you can see the power of this. But in the case of gravity, this is what it means. So, this agrees with the intuitive notion that when you lose height you gain speed, when you lower your _{2}y you gain speed, but it’s much more precise than that. Instead of saying vaguely, “you lose and gain,” ½mv is the part connected with the speed and ^{2}mgy is the part connected with the height and that sum does not change with time.So, let’s go back to the mass and spring system. In the case of the mass and spring system, if you have a mass connected to a spring, then we say here ½ kx equal to ^{2}E is a constant. That means, find it at any initial time one, find it at any final time two, add these two numbers, the total will not change.So, let me put that to work. Let me use that in a calculation. So, I say here is a spring; here is where it likes to be. I’m going to pull it to a new location If you go back to Newton’s laws, it’s a pretty complicated problem. You have got to think about why it’s a complicated problem. Because you start with a mass at rest, if you pull it by an amount In fact, let’s do it not just to be at the mid point here, but at any point kx = ½_{1}²m v + ½_{2}^{2}kx. This is the truth; this is a constant that does not change with time. Now, to get some mileage out of this, you’ve got to know what the constant value is. Well, we know what it is in the beginning because when you pull this guy, at the instant we knew it had no velocity, and we knew it was sitting at the point _{2}²A, so we know ½k A is the total energy of this mass and spring system. That’s not going to change with time. We like to say that at the initial time all the energy is potential energy; there is no kinetic energy because there is no motion. At any subsequent time, this number has to be equal to what you get. Now, let me drop the subscript to there [pointing to equation]. Let it stand for some generic instant in the future. This much we know. Now, do you see that we can solve for the velocity on any location _{1}²x? You pick the x; if you want x to be zero that’s very easy, you kill that term, you balance these two, you cancel the half and you can see that’s a formula for v. ^{2}v is going to be ^{2}ka over ^{2}m; that is going to be the velocity when it swings by this midpoint. But you can pick any x you like, if the x was not zero but .1; well, put the .1, take it to the other side and solve for v. This is one example of the Law of Conservation of Energy.So, the power of the Law of Conservation of Energy is if you knew the initial energy, you don’t have to go through Newton’s laws, and you’ll find it in all the problem sets and many exams. When a problem is given to you and you are told to find the velocity of something, you should try whenever possible to use the Law of Conservation of Energy rather than going to Newton’s laws. So, if you want to know what have we done, how come finding the velocity seemed so difficult when I described it to you earlier, and would we have done it, the answer is, if you tell me the force of the spring is a function of
U will be -x/2. So, the force is ^{2}-kx, then you have to find the U so that -du/dx = -kx and the answer to that is U is kx/2. So, if you knew the force, you just find the function ^{2}U by asking, “Whose derivative gives me F with a minus sign?” Then you are done. That’s the potential energy for that problem.Now, lets take another problem, a mass is hanging from the ceiling. Then, it’s got two kinds of potential energy, one is just a gravitational potential energy, and the other is a spring potential energy, and that’s going to come because F(x)dx. I’m sorry, let’s call it dy and F(y)dy from y to _{1}y. And this _{2}F(y) will have two parts, the force of gravity – I’m tired of writing the limits – then, the force of the spring, from 1 to 2 [adding the limits to the integrals]. Right, there are two forces acting on it. This one will be mgy, this one will be ½_{2} - mgy_{1}ky - ½_{2}^{2}ky. So, every force acting on the body will turn into a potential energy, because the force–Newton’s law says Work Energy Theorem applies to the total force. So, the total force is two parts, and you can do two integrals; each integral is an upper limit minus lower limit of some function and this is what you get [pointing to formula]._{1}^{2}Oh, I think I made one mistake here. I think this should give me, yes, this overall sign is wrong in these equations. Really, there’s a minus in that and minus in that. In other words, the work done by the force, if there’s only one force acting, is really G and U are minuses of each other. So, integral of force will give U initially -U finally. So, it should really be + mgy._{1} - mgy_{2}So, the Law of Conservation of Energy for this object will be that, ½ ky is the total energy and will not change. So, this mass can bump up and down and go back and forth. If you knew the total of these three numbers at one instant, you know the value at any other instant. For example, if you pulled it from its equilibrium by .2 centimeters and you released it, and you say what’s the speed when it’s .1 centimeters away, it’s very easy because you know the initial energy was all spring energy and gravitational energy. At any later time if you knew the ^{2}y, if you knew the position, you know the potential energy; that’s the whole point. Potential energy depends on where you are; kinetic energy depends on how fast you’re moving. And some combination of where you are and how fast you’re moving is invariant, does not change with time. So, if you tell me where you are, I will tell you how fast you’re moving. By the way, remember the kinetic energy enters as v. You can imagine, when you solve for ^{2}v you’re always going to get two answers. Why is that? Does it make any sense? Yes?
## Chapter 4. Friction Force Effect on Work-Energy Theorem [00:44:39]Okay, now, we are going to find one bad boy, bad girl, bad person, which is going to ruin this whole thing. There is–you know what I’m talking about here, yes?
G(x or which is _{2}) - G(x_{1})U(x; you can rearrange it to get _{1}) - U(x_{2})K _{2} + U_{2}= K. Looks like you can always get Law of Conservation of Energy. And you want to know why that fails when there is friction._{1} + U_{1}Okay, why does it fail when there is friction? So, when there is friction–Let me try to do it again. -kx. Let me do that integral plus the force of friction dx, from x to _{1}x. This one is from _{2}x to _{1}x and we saw what this is. Let me write it one more time, ½_{2}kx makes ½_{1}^{2}kx; if you didn’t have friction you are done. Because you can get kinetic plus potential for spring, is kinetic plus potential. Why don’t we just integrate the force of friction and then turn that into a _{2}^{2}U and then we’re done? What’s going to be the problem? Let me just give some–anybody else want–yes?
Therefore, we do this integral but we don’t do this. Even though I write it this way, mathematically it’s wrong because the force of friction–Is it really a function of U, so you’ll get _{2}(K is equal to this last thing, work done by friction from _{2} + U_{2}) - (K_{1}+U_{1})x to _{1}x and the question is, “Can I do the integral?” The answer is, if you just tell me _{2}x and _{1}x, then I know that’s the point you’re making; I cannot just do an integral because I will have to know the whole story, but in limited cases I can do this. Suppose I pulled a mass by an amount _{2}A and I let it go, but now there is friction. I think we all know that if it came to the center it won’t be going quite as fast. And we also know, when it overshoots the other side, it will not go back to -A; normally, it will go back to -A. Everyone knows why it goes back to -A, right? When you go the other extremity without friction, your energy is all potential. It was ½kA to begin with, it’s going to be ½^{2}kA at the end, except ^{2}A can go to -A without changing the answer; that’s why it’ll go from A to -A. But when you’ve got friction, I cannot say the energies are equal because I’ve got to do this integral. For this part of the journey from here to here, during that part of the journey, the force is a well-defined function. Namely, it is a constant value -µ times mg and it is acting this way. So, for this part of the trip, I can do the integral. It is very easy to find out what it is; it’s -mg times µ kinetic, times the distance traveled.Of course, it’s a little more complicated now. You have to tell me where you want your velocities. You might say, find the velocity at mv which is what I’m trying to find. I am at the origin so it’s ½^{2}k times 0^{2}, no potential energy. Initially, there is no kinetic energy but there was potential energy ½kA. You follow that logic now? This is the difference in energies due to the conserved forces, the conserved energies, but the difference ^{2}E is no longer 0 but is given by this number. So, I can calculate it for this part._{2} - E_{1}I can even ask, – this is one of the homework problems, I think – you’re asked, “How far does it go to the left?” Well, what do you do? Let’s call the place where it goes to. The left most extremity is A to come here and another A′ to go to the left. A′ is a magnitude of how far you go to the left. And likewise, on the left-hand side you’ve got to have the total kinetic plus potential minus kinetic plus potential. At the end of the day, you only have potential energy ½k A′ squared, no kinetic and initially, again, you only have potential energy ½kA. You will see that A prime now satisfies a quadratic equation; you can solve it.^{2}But what is the point? The point is you cannot do this integral once and for all and say the answer depends on the end points. In this particular problem, if you told me you are doing this part of the journey, during that part of the journey because for the entire trip, the velocity had a particular sign, namely to the left, force of friction had a definite magnitude and sign, namely to the right, and then of course it’s as good as any legitimate function and you can do the integral. In fact, the forces are constant; you can pull it out of the integral, and the integral Okay, so the final bottom line for Law of Conservation of Energy that you guys should remember is the following. You take all the forces acting on a body. So, lets do it one more time in our heads, so you know where everything comes from. Everything comes from this result F, and one is a force of friction, _{g} dxdx. This guy will turn into a potential energy difference. Let me just take one good force and the force of friction. If you had two, each will generate its own potential when you do the integral. Then, we may then say E; that’s the same as _{2} - E_{1}(K is equal to the work done by friction. Work done by friction is the shorthand for that integral which you can do only; you must divide the motion involved into pieces. For example, if you have a mass that swung to the left then turns around and goes back to the right, of course, again, it won’t come back to the original, it’ll come to a smaller number. That’s because there’s more loss of energy on the return trip. For the return trip, you must do the problem separately. On the return trip, you start from this _{2} + U_{2}) - (K_{1} +U_{1})A′ that you found, then start moving to the right, the friction will act to the left. So, you must divide the motion into segments. During each segment the force should have a definite direction and magnitude so you can do the integral. The point is, there is no universal formula for F(x) due to friction.So, in one dimension that’s all it is, bringing the good and bad forces, those which lead to potential, those which don’t; the difference is the good forces are functions of ## Chapter 5. Calculus Review: Small Changes [00:56:13]Okay? Alright now, what I have to do here for the remaining time is a little more of some mathematical preparation for what I’m going to do next. What I’m going to do next is to go to higher dimensions and we’ve got to make sure we all know some of the mathematical ideas connected to the higher dimensions. So, I’m going to start off with some things and this is like a mathematical interlude. Again, I told you, if you’re not strong on some of these things you should read this math book I mentioned to you on So, let’s take some function So, let’s take a concrete example. Suppose F(x) = x x Δx + Δx. Therefore, the change in ^{2}F – you got to take it to the other side – change in F = 2x Δx + Δx. 2^{2}x Δx, 2x is, of course, what you realize is that the derivative of the function at that point. And that’s giving you this part but there’s a tiny portion on the top, which is Δx that you’re not getting. If the function is ^{2}x, then ^{3}x to the fourth, then hyperbolic cos (x) there’ll be more errors. But the point is, the change in the function has a leading piece that’s proportional Δx, and then the corrections that we see mathematically are of higher order in Δx.So, if So, let’s take an example of this function, of this rule, because I’m going to use it all the time. So, one example is
x. But you have evaluated it at x = 0, so this goes away, then you get nx. So, the change in the function, this function as you move off x = 0 is nx. So, what we are saying then is 1 + x raised to the n is 1 + nx + tiny numbers, if x is small. That is a very useful result. If something looks like 1 + x raised to a power, and if you’re not going too far from the point 1, namely if your x is very small, then don’t bother with doing the whole 1 + x times 1 + xn times, it’s approximately equal to 1 + nx.So, let me give you one example; you will use this later on. According to relativity, a mass of a particle is given by this [writes formula on the board]. It’s a body sitting still, it has certain mass. In Newtonian mechanics we say the mass is the mass and that’s the end of it. Relativity, you don’t have to know the origin of this; that’ll be taught to you in due course. But it just says particle to the velocity have a mass different from when they are at rest. And this is the velocity and Now, most of the time we don’t care about it because v over ^{2}c to the power of -½ , do you agree? That’s what a square root in the bottom is, it’s this power upstairs. Now, we can write it as follows, ^{2}m, this is (1 + x)_{0}^{n}, okay? So the x that we use is -v and the ^{2}/c^{2}n that we use is -½. That’s what I’m using in this formula; x is -v/^{2}c and ^{2}n, which is the power to which I’m raising, is -½; therefore, this is equal to 1 + v/2^{2}c plus corrections, which are negligible if ^{2}v is very small. This is a very useful result. It tells you the mass variation is that is the mass at rest plus ^{2}/c^{2}m2_{0}v^{2}/c, it’s a very useful formula. Sometimes this formula is much more friendly to use than the exact formula. You cannot beat the exact formula; it’s the exact truth but if you’re interested in that exact function for small values of ^{2}v/c, it’s much more useful to use this approximation. And you will have a lot of problems like this in your homework, now, later, everywhere else. So, if I’m going to use it all the time I should tell you what it is.So, let me summarize what I’ve told you so far. First result is, if a function is known at a certain point and you want to know its value at a nearby point, what are you supposed to do? That’s not so hard to guess right? For example, your tuition costs are so much this year at Yale. And we know how much it’s going up every year; you can make a guess on what it’ll be a year from now. Of course in reality, it’ll be higher than that because there is other terms, but that’s the best guess you would make. That’s how we predict the future by looking at past rates of change; that’s very normal. The point is that you can predict the tuition in a year from now maybe, but maybe not five years from now or ten years from now because there’s inflation on top of inflation. That’s like saying the rate of change itself has a rate of change, the slope here is not the same as the slope there, that’s what the higher derivatives of the function do for you. So, I will conclude by telling you one last result which I’m not going to prove in this course. And you guys should learn from the math department the following result. There is some crazy function, okay, this is But of course, d, ^{3}xf/dx times ^{3}x. This goes on forever. And it says that if you knew every derivative of the function at the origin you can reconstruct the function. Not just infinitesimally close to it, but as far as you want to go, there are some mathematical restrictions on this because there are infinite number of terms in this thing. When you add them all up you may get infinity which would be nonsense. So, you should really sum up this infinite number of terms if they add up to a finite number, and if they do that will give you the function here. What we have done is to just use the first two terms as an approximation. Or what I’m telling you is that there is something called The Taylor Expansion of a function that allows you to know the function even here, macroscopically, far from here if you knew all the derivatives. And if you put those derivatives here and if the sum added up to something finite, that finite thing is the value of the function at the new point.^{3}Here is an example you guys can take, f(x) at all points? You have got to remember that e has a beauty, that the derivative is ^{x}e itself. And its value at the origin of every derivative is simply one. So, this just becomes 1 + ^{x}x + x/2. And that is, in fact, the expression for ^{2}e and it turns out that no matter how big ^{x}x is, the sum in fact adds up to a finite number that we call e. And in particular, what is ^{x}e itself, e itself e to the 1; e is e to the 1 so put x = 1 you get 1 + 1 + 1^{2} over 2 factorial plus 1^{3} over 3 factorial. You do all that you get 2 point something, something. Alright, that’s e. So, this is a very powerful notion. If you knew all the derivatives of a function, you can predict its value not only the tiny neighborhood, but over big distance. If you knew the first derivative and you were not very ambitious. Look, the logic is very simple. If x is a tiny number x is even tinier, ^{2}x is even tinier. We’ve got no respect for these terms. We just drop them and we stop here. Okay, I’ll have something more to say about this later on. But for now, these ideas will be useful for what we’re doing next time.^{3}[end of transcript] Back to Top |
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